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Question

If sin1x+sin1y+sin1z=π, prove that x1x2+y1y2+z1z2=2xyz.

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Solution

We are given,
sin1x+sin1y+sin1z=π.

let,
sin1x=AsinA=x
sin1y=BsinB=y
sin1z=CsinC=z.

here,
A,B,C[π/2,π/2]
A+B+C=π ........ (1)

now,
LHS =x1x2+y1y2+z1z2

=sinA.1sin2A+sinB1sin2B+sinC.1sin2C

=sinA.cosA+sinBcosB+sinC.cosC

[Multiply & Divide by 2]

=22[sinAcosA+sinBcosB]+sinC.cosC

=[2sinAcosA+2sinBcosB]2+sinC.cosC(2sinθcosθ=sin2θ)

=(sinA+sin2B)2+sinC.cosC

=2sin(2A+2B2)cos(2A2B2)2+sinC.cosC

=sin(A+B)cos(AB)+sinC.cosC

=sin(πC)cos(AB)+sinC.cos(π(A+B))

=sinC[cos(AB)cos(A+B)]

=sinC[2sin(2B2)sin(3A2)]

=2sinAsinBsinC

=2xyz

=RHS.

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